A truck pulls boxes up an incline plane. The truck can exert a maximum force of #3,200 N#. If the plane's incline is #(5 pi )/8 # and the coefficient of friction is #3/7 #, what is the maximum mass that can be pulled up at one time?

Question

A truck pulls boxes up an incline plane.  The truck can exert a maximum force of #3,200 N#.  If the plane's incline is #(5 pi )/8  # and the coefficient of friction is #3/7  #, what is the maximum mass that can be pulled up at one time?

Emily Abbate · Accepted Answer

#m_"max" = 322# #"kg"#

We're asked to find the maximum mass a truck can pull up at one time, with some given information.

We'll treat the boxes as one body, and our goal is finding the mass of this body, #m#.

The maximum mass the truck can pull is the largest so that the net force is not down the incline (then the boxes would not be going up, intuitively). Thus, at this condition, the boxes are in equilibrium (no net forces acting, pulls upward with constant velocity).

Thus, treating the positive #x#-axis as up the incline:

#sumF_x = overbrace(F_"truck")^(3200color(white)(l)"N") - overbrace(f_k)^(= mu_kn = 3/7mgcostheta) - mgsintheta = 0#

#3200# #"N"# #- 3/7mgcostheta - mgsintheta = 0#

#3200# #"N"# #= 3/7mgcostheta + mgsintheta#

Divide all terms by the mass, #m#:

#(3200color(white)(l)"N")/m = 3/7gcostheta +gsintheta#

Therefore,

#(3200color(white)(l)"N")/(3/7gcostheta +gsintheta) = m#

#= (3200color(white)(l)"N")/(3/7(9.81color(white)(l)"m/s"^2)sin((5pi)/8) + (9.81color(white)(l)"m/s"^2)cos((5pi)/8))#

#= color(red)(322# #color(red)("kg"#

Lily Adams · Answer

undefined The maximum mass that can be pulled up the incline can be calculated using the formula:

$$ m = \frac{F_{\text{max}} - F_{\text{friction}}}{g \sin(\theta) + \mu_k g \cos(\theta)} $$

where:
- $ F_{\text{max}} $ is the maximum force the truck can exert (3,200 N),
- $ F_{\text{friction}} $ is the force of friction ($ \mu_k \cdot F_{\text{normal}} $),
- $ \theta $ is the angle of the incline (5π/8),
- $ \mu_k $ is the coefficient of friction (3/7),
- $ g $ is the acceleration due to gravity (9.8 m/s²).

Plugging in the given values:

$$ m = \frac{3200 - \frac{3}{7} \cdot mg \cos\left(\frac{5\pi}{8}\right)}{g \sin\left(\frac{5\pi}{8}\right) + \frac{3}{7} \cdot g \cos\left(\frac{5\pi}{8}\right)} $$

$$ m = \frac{3200 - \frac{3}{7} \cdot (9.8) \cdot m \cdot \cos\left(\frac{5\pi}{8}\right)}{(9.8) \cdot \sin\left(\frac{5\pi}{8}\right) + \frac{3}{7} \cdot (9.8) \cdot \cos\left(\frac{5\pi}{8}\right)} $$

$$ m = \frac{3200 - \frac{29.4}{7} \cdot m \cdot \cos\left(\frac{5\pi}{8}\right)}{9.8 \cdot \sin\left(\frac{5\pi}{8}\right) + \frac{29.4}{7} \cdot \cos\left(\frac{5\pi}{8}\right)} $$

Solving for $ m $:

$$ m = \frac{3200}{9.8 \cdot \sin\left(\frac{5\pi}{8}\right) + \frac{29.4}{7} \cdot \cos\left(\frac{5\pi}{8}\right) + \frac{29.4}{7}} $$

$$ m \approx \frac{3200}{9.8 \cdot 0.9239 + \frac{29.4}{7} \cdot (-0.3827) + \frac{29.4}{7}} $$

$$ m \approx \frac{3200}{9.049 + (-3.059) + 4.2} $$

$$ m \approx \frac{3200}{10.19} $$

$$ m \approx 314.37 \, \text{kg} $$

Therefore, the maximum mass that can be pulled up at one time is approximately 314.37 kg.